Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A at $(45,10)$, B at $(10,20)$, $AB=AC$ and angle $C=20$ degree find the coordinates of $C$.suggest the formula so i can write code in Perl.

share|improve this question
add comment

2 Answers 2

You have the coordinates of A and B, so you can compute the distance AB. AB=AC, so you then know the distance AC. Let the coordinates of C be (x,y). Apply the distance formula to A and C and set the result equal to the distance you already computed. This equation guarantees that AB=AC.

Now, the angle at C is determined by the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$. These vectors can be found by subtracting the coordinates of C from A and B (respectively). $(CA)(CB)\cos C = \overrightarrow{CA}\cdot\overrightarrow{CB}$, and you know the distances CA = AC = AB and the measure of C, and you can compute CB and the dot product in terms of (x,y). This equation guarantees that the measure of C is 20°.

Solve the system resulting from the two equations above to find the coordinates of C. There are almost certainly 2 solutions.

(See also this question.)

share|improve this answer
add comment

We have $A(45,10),B(10,20),C(x_c,y_c)$.

$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$

$C=B=20^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=\pi /9$ rad

$A=180^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }-40^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=140^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=7\pi /9$ rad

alt text

Let's make the following change of variables: $x=X+45,y=Y+10$ (translation of axes). Then $A$ becomes the origin of the $XY$ referential.

The vector $\overrightarrow{AB}$ can be written in this $XY$ referential as

$\overrightarrow{AB}=(5\sqrt{53}\cos \left( \pi -\arctan \frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\arctan \frac{2}{7}\right) )=(-35,10)$

and the vector $\overrightarrow{AC}$ as

$\overrightarrow{AC}=(5\sqrt{53}\cos \left( \pi -\frac{7\pi }{9}-\arctan \frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\frac{7\pi }{9}-\arctan \frac{% 2}{7}\right) )$

Therefore in the original referencial $xy$, we have

$x_{C}=5\sqrt{53}\cos \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi \right) +45\approx 78.239$

$y_{C}=5\sqrt{53}\sin \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi \right) +10\approx 24.837$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.